使用LINQ Concat List <string>中的所有字符串

Question

是否有任何简单的LINQ表达式将我的整个List<string>集合项连接到string具有分隔符的单个？

如果集合是自定义对象而不是string？想象一下,我需要连接object.Name.

Answer 1

在.NET 4.0或更高版本中:

String.Join(delimiter, list);

足够了.

Answer 2

通过使用LINQ,这应该工作;

string delimiter = ",";
List<string> items = new List<string>() { "foo", "boo", "john", "doe" };
Console.WriteLine(items.Aggregate((i, j) => i + delimiter + j));

课程描述:

public class Foo
{
    public string Boo { get; set; }
}

用法:

class Program
{
    static void Main(string[] args)
    {
        string delimiter = ",";
        List<Foo> items = new List<Foo>() { new Foo { Boo = "ABC" }, new Foo { Boo = "DEF" },
            new Foo { Boo = "GHI" }, new Foo { Boo = "JKL" } };

        Console.WriteLine(items.Aggregate((i, j) => new Foo{Boo = (i.Boo + delimiter + j.Boo)}).Boo);
        Console.ReadKey();

    }
}

这是我最好的:)

items.Select(i => i.Boo).Aggregate((i, j) => i + delimiter + j)

Answer 3

这是一个字符串数组:

string.Join(delimiter, array);

这是一个List <string>:

string.Join(delimiter, list.ToArray());

这是一个自定义对象列表:

string.Join(delimiter, list.Select(i => i.Boo).ToArray());

Answer 4

using System.Linq;

public class Person
{
  string FirstName { get; set; }
  string LastName { get; set; }
}

List<Person> persons = new List<Person>();

string listOfPersons = string.Join(",", persons.Select(p => p.FirstName));

Answer 5

好问题.我一直在用

List<string> myStrings = new List<string>{ "ours", "mine", "yours"};
string joinedString = string.Join(", ", myStrings.ToArray());

它不是LINQ,但它有效.

Answer 6

您可以简单地使用：

List<string> items = new List<string>() { "foo", "boo", "john", "doe" };

Console.WriteLine(string.Join(",", items));

快乐编码！

Answer 7

List<string> strings = new List<string>() { "ABC", "DEF", "GHI" };
string s = strings.Aggregate((a, b) => a + ',' + b);

Answer 8

我认为如果你在扩展方法中定义逻辑,代码将更具可读性:

public static class EnumerableExtensions { 
  public static string Join<T>(this IEnumerable<T> self, string separator) {  
    return String.Join(separator, self.Select(e => e.ToString()).ToArray()); 
  } 
} 

public class Person {  
  public string FirstName { get; set; }  
  public string LastName { get; set; }  
  public override string ToString() {
    return string.Format("{0} {1}", FirstName, LastName);
  }
}  

// ...

List<Person> people = new List<Person>();
// ...
string fullNames = people.Join(", ");
string lastNames = people.Select(p => p.LastName).Join(", ");